Question
If $a - b = 0.9$ and $ab = 0.36;$ find:$(i) \ a + b;(ii) \ a^2- b^2.$
✓
Answer
$(i)$ We know that,
$(a-b)^2=a^2-2 a b+b^2$ and $(a+b)^2=a^2+2 a b+b^2$
Rewrite the above equation, we have
$(a+b)^2 =a^2+b^2-2 a b+4 a b$
$ =(a-b)^2+4 a b$
Given that $a-b=0.9 ; a b=0.36$
Substitute the values of $(a-b)$ and $(ab)$
in equation $(1), $we have
$(a+b)^2 =(0.9)^2+4(0.36)$
$ =0.81+1.44=2.25$
$\Rightarrow a+b = \pm \sqrt{2.25}$
$\Rightarrow a+b = \pm 1.5$
$(ii)$ We know that,
$a^2-b^2=(a+b)(a-b)$
From equation $(2)$ we have,
$a+b= \pm 1.5$
Thus equation $(3)$ becomes,
$a^2-b^2=( \pm 1.5)(0.9) [$given $a-b=0.9]$
$\Rightarrow a^2-b^2= \pm 1.35 $
$(a-b)^2=a^2-2 a b+b^2$ and $(a+b)^2=a^2+2 a b+b^2$
Rewrite the above equation, we have
$(a+b)^2 =a^2+b^2-2 a b+4 a b$
$ =(a-b)^2+4 a b$
Given that $a-b=0.9 ; a b=0.36$
Substitute the values of $(a-b)$ and $(ab)$
in equation $(1), $we have
$(a+b)^2 =(0.9)^2+4(0.36)$
$ =0.81+1.44=2.25$
$\Rightarrow a+b = \pm \sqrt{2.25}$
$\Rightarrow a+b = \pm 1.5$
$(ii)$ We know that,
$a^2-b^2=(a+b)(a-b)$
From equation $(2)$ we have,
$a+b= \pm 1.5$
Thus equation $(3)$ becomes,
$a^2-b^2=( \pm 1.5)(0.9) [$given $a-b=0.9]$
$\Rightarrow a^2-b^2= \pm 1.35 $
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